Solve (a^2+1)(c^2+1)=(ac-1)^2+(a+c)^2

$$( a ^ { 2 } + 1 ) ( c ^ { 2 } + 1 ) = ( a c - 1 ) ^ { 2 } + ( a + c ) ^ { 2 }$$

Solve for a

$$a\in \mathrm{R}$$

Solution Steps

Use the distributive property to multiply a^{2}+1 by c^{2}+1.

$$a^{2}c^{2}+a^{2}+c^{2}+1=\left(ac-1\right)^{2}+\left(a+c\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(ac-1\right)^{2}.

$$a^{2}c^{2}+a^{2}+c^{2}+1=a^{2}c^{2}-2ac+1+\left(a+c\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a+c\right)^{2}.

$$a^{2}c^{2}+a^{2}+c^{2}+1=a^{2}c^{2}-2ac+1+a^{2}+2ac+c^{2}$$

Combine -2ac and 2ac to get 0.

$$a^{2}c^{2}+a^{2}+c^{2}+1=a^{2}c^{2}+1+a^{2}+c^{2}$$

Subtract a^{2}c^{2} from both sides.

$$a^{2}c^{2}+a^{2}+c^{2}+1-a^{2}c^{2}=1+a^{2}+c^{2}$$

Combine a^{2}c^{2} and -a^{2}c^{2} to get 0.

$$a^{2}+c^{2}+1=1+a^{2}+c^{2}$$

Subtract a^{2} from both sides.

$$a^{2}+c^{2}+1-a^{2}=1+c^{2}$$

Combine a^{2} and -a^{2} to get 0.

$$c^{2}+1=1+c^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{R}$$

Solve for c

$$c\in \mathrm{R}$$

Solution Steps

Use the distributive property to multiply a^{2}+1 by c^{2}+1.

$$a^{2}c^{2}+a^{2}+c^{2}+1=\left(ac-1\right)^{2}+\left(a+c\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(ac-1\right)^{2}.

$$a^{2}c^{2}+a^{2}+c^{2}+1=a^{2}c^{2}-2ac+1+\left(a+c\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+c\right)^{2}.

$$a^{2}c^{2}+a^{2}+c^{2}+1=a^{2}c^{2}-2ac+1+a^{2}+2ac+c^{2}$$

Combine -2ac and 2ac to get 0.

$$a^{2}c^{2}+a^{2}+c^{2}+1=a^{2}c^{2}+1+a^{2}+c^{2}$$

Subtract a^{2}c^{2} from both sides.

$$a^{2}c^{2}+a^{2}+c^{2}+1-a^{2}c^{2}=1+a^{2}+c^{2}$$

Combine a^{2}c^{2} and -a^{2}c^{2} to get 0.

$$a^{2}+c^{2}+1=1+a^{2}+c^{2}$$

Subtract c^{2} from both sides.

$$a^{2}+c^{2}+1-c^{2}=1+a^{2}$$

Combine c^{2} and -c^{2} to get 0.

$$a^{2}+1=1+a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any c.

$$c\in \mathrm{R}$$